maximum value of numerical pattern

Hi Everyone,

I have a problem that I'm slightly struggling with.

I have two arrays which represent the time step (A) and the corresponding iterations at each time step (B).

A = [1 1 1 2 2 2 3 3 3 3 4 4 4 4 4]
B = [1 2 3 1 2 3 1 2 3 4 1 2 3 4 5]

I would like to extract the maximum iteration (from B) for each time step (from A).

So it should look like

C = [1 2 3 4]
D = [3 3 4 5]

Any help is greatly appreciated.

Best regards,

Lloyd

Answer 1

[Stephen23]

>> A = [1 1 1 2 2 2 3 3 3 3 4 4 4 4 4];
>> B = [1 2 3 1 2 3 1 2 3 4 1 2 3 4 5];
>> C = unique(A)
C =
   1   2   3   4
>> D = accumarray(A(:),B(:),[],@max)
D =
   3
   3
   4
   5

[Lloyd Sullivan]

Thank you for the prompt response Stephen, it works perfectly! I never knew of this accumarray command, I have learnt something new :)

[Steven Lord]

That works as long as the elements in A and B are positive integer values. Using findgroups and splitapply works when A and B contain positive integer values:

A = [1 1 1 2 2 2 3 3 3 3 4 4 4 4 4];
B = [1 2 3 1 2 3 1 2 3 4 1 2 3 4 5];
G = findgroups(A);
results = splitapply(@max, B, G)

It also works if they contain something other than positive integer values:

A2 = A./4;
G2 = findgroups(A2);
results = splitapply(@max, B, G2)

Operating on A2 using accumarray will throw an error:

D = accumarray(A2(:), B(:), [], @max)

[Image Analyst]

And Steve's answer does not work in the general case where the numbers in B can be anything, like even negative numbers. Try this and see:

A = [1 1 1 2 2 2 3 3 3 3 4 4 4 4 4];
B = rand(1, length(A)) - 0.5
G = findgroups(A);
results = splitapply(@max, B, G)

If you want something what works for any values of A and any values of B, regardless of sorting order or positive of negative values, it's a few more lines of code. Hint: one way is to use regionprops() in the Image Processing Toolbox.

[Steven Lord]

Image Analyst, to which of us were you referring when you said "Steve's answer"? If it was to me, the code you posted works when I try it in release R2018a.

>> rng default
>> A = [1 1 1 2 2 2 3 3 3 3 4 4 4 4 4];
>> B = rand(1, length(A)) - 0.5;
>> G = findgroups(A);
>> results = splitapply(@max, B, G);
>> [A.', B.'], results

ans =

      1.0000    0.3147
      1.0000    0.4058
      1.0000   -0.3730
      2.0000    0.4134
      2.0000    0.1324
      2.0000   -0.4025
      3.0000   -0.2215
      3.0000    0.0469
      3.0000    0.4575
      3.0000    0.4649
      4.0000   -0.3424
      4.0000    0.4706
      4.0000    0.4572
      4.0000   -0.0146
      4.0000    0.3003

results =

      0.4058    0.4134    0.4649    0.4706

Do you see different answers when you run those six lines of code? If so what release are you using?

Answer 2

[Adam Danz]

Differentiate B in order to know when the count restarts. Then use the index of the restart-points to get C and D.

maxIdx = diff([B,-1])<0;    % -1 needed to capture last value
C = A(maxIdx); 
D = B(maxIdx);

[Adam Danz]

Here's an alternative solution if speed is an issue (both of these solutions are fast but this one is faster).

[Lloyd Sullivan]

Very efficient - almost twice as fast. Thanks you Adam.